Find the locus of the point R on L such that the distances BP,BR and BQ are in harmonic progression.

Question

A variable line L passing through the point $B(2,5)$ intersect the lines $2x^2-5xy+2y^2=0$ at P and Q.Find the locus of the point R on L such that the distances BP,BR and BQ are in harmonic progression.

I decomposed the pair of straight lines into two lines.But after that I dont know what to do....

Answer 1

Let $R(h, k)$ be the coordinates of the point $R$ then the equation of the line $L$ passing through the points $B(2, 5)$ & $R(h, k)$ is given as $$y-k=\frac{k-5}{h-2}(x-h)$$ $$y=\frac{k-5}{h-2}x+\frac{5h-2k}{h-2}$$

Now, the lines represented by $2x^2-5xy+2y^2=0$ or $(2x-y)(x-2y)=0$ passing through the origin given as $$y=2x, \ y=\frac 12x$$ Now, the intersection point of lines: $y=2x$ & $L$is $P\left(\frac{5h-2k}{2h-k+1},\frac{10h-4k}{2h-k+1}\right)$ & similarly, the intersection point of lines: $y=\frac 12x$ & $L$ is $Q\left(\frac{10h-4k}{h-2k+8},\frac{5h-2k}{h-2k+8}\right)$

Now, using distance formula, one should have
$$BR=\sqrt{(h-2)^2+(k-5)^2}$$ $$BP=\sqrt{\left(\frac{5h-2k}{2h-k+1}-2\right)^2+\left(\frac{10h-4k}{2h-k+1}-5\right)^2}=\frac{\sqrt{(h-2)^2+(k-5)^2}}{2h-k+1}=\frac{BR}{2h-k+1}$$ $$BQ=\sqrt{\left(\frac{10h-4k}{h-2k+8}-2\right)^2+\left(\frac{5h-2k}{h-2k+8}-5\right)^2}=\frac{8\sqrt{(h-2)^2+(k-5)^2}}{h-2k+8}=\frac{8BR}{h-2k+8}$$ Since, $BP, BR, BQ$ are H.P. hence $$BR=\frac{2(BP)(BQ)}{BP+BQ}$$ Now, setting the values, one should get $$BR=\frac{2\frac{BR}{2h-k+1}\frac{8BR}{h-2k+8}}{\frac{BR}{2h-k+1}+\frac{8BR}{h-2k+8}}$$ $$1=\frac{16}{h-2k+8+16h-8k+8}$$ $$17h-10k=0$$ substituting $h=x$ & $k=y$ one should get the locus of point $R$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{17x-10y=0}}$$

Answer 2

The variable line L can be given by (here the slope $m$ is the variable) $$y=m(x-2)+5.$$ The pair of lines given are $$x-2y=0 \qquad \text{ and } \qquad y-2x=0.$$

Now find the intersection points P and Q. Once you have those (for example P will be $\left(\frac{2(2m-5)}{2m-1}, \frac{2m-5}{2m-1}\right)$) then you can compute the distances BP, BQ. Now take point R as $(h,k)$. Compute the distances BR. With the harmonic condition, you will be able to eliminate $m$ and get an equation in terms of $h$ and $k$. That will be your locus.

Let me know if you need something to be more elaborate.